Thread: Finding Ker(f)

Hello !
I'm a bit confused about the notion of Ker() and the solution I was given for an exercise.

f is a linear application defined for u = (x,y,z) by :

f(u) = (x + y + z, x + 2y + 3z, x + 3y + 5z)

Written as a system of equations :

$\displaystyle \left\{\begin{matrix} x + y + z = 0
\\ x + 2y + 3z = 0
\\ x + 3y + 5z = 0

\end{matrix}\right.
$

I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)

However, I have no idea how to get to this vector.
I solved the system by putting it in a rref matrix and got as solution x = z, y = -2z.
So this would give me v1 = (1,-2,1) if z equals one ? So I get Ker(f) = 1 also but I don't get why my vector is different and not even collinear to the one from the solution.

I also have trouble understanding how to interpret what Ker() or Im() represent (geometrically or just understand the concept).

Thank you for your help !

The image of a function is the set of all possible values the function can take, which is a subset of the function's codomain. The kernel of a function is the set of all possible inputs to a function that the function will take to zero, so the kernel is a subset of the domain of the function.

Your vector is correct. Try plugging in the vector you were given as a solution, and you will find that it does not yield the zero vector.

Originally Posted by Jo37

Hello !
I'm a bit confused about the notion of Ker() and the solution I was given for an exercise.

f is a linear application defined for u = (x,y,z) by :

f(u) = (x + y + z, x + 2y + 3z, x + 3y + 5z)

Written as a system of equations :

$\displaystyle \left\{\begin{matrix} x + y + z = 0\\ x + 2y + 3z = 0\\ x + 3y + 5z = 0\end{matrix}\right.$

I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)

However, I have no idea how to get to this vector.
I solved the system by putting it in a rref matrix and got as solution x = z, y = -2z.
So this would give me v1 = (1,-2,1) if z equals one ? So I get Ker(f) = 1 also but I don't get why my vector is different and not even collinear to the one from the solution.
I also have trouble understanding how to interpret what Ker() or Im() represent (geometrically or just understand the concept).

Look at this.
Null space is another name for kernel.

Originally Posted by Jo37

I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)

I have no idea what this means! If v1 is a vector what does "Vect{v1}" mean? The vector space spanned by v1? But then what does "Vect{v1}= 1" mean? How is a vector space equal to 1? In any case, "(-2, 3, 1)" is not in the kernel. In particular, x+ y+ z= -2+ 3+ 1= 2, not 0!

Given x + y + z = 0, x + 2y + 3z = 0, x + 3y + 5z = 0, we can subtract the first equation from the second to get y+ 2z= 0 and subtract the first equation from the third to get 2y+ 4z= 0. From the first of those, y= -2z so 2y= -4z and the last equation becomes -4z= -4z, true for all z. With y= -2z, x+ y+ z= x- 2z+ z= x- z= 0 so x= z. That is, (x, y, z)= (z, -2z, z) satisfies all three equations for any number, z. We can write that as (z, -2z, z)= z(1, -2, 1). The kernel is the one dimensional subspace of all multiples of (1, -2, 1).

Thank you very much for your answers !

@HallsofIvy This is literally what was written in the solution I was given. How to write the solution properly ? Ker(f) = (z, -2z, z) and dim(Ker) = 1 ?

Originally Posted by Jo37

Thank you very much for your answers !

@HallsofIvy This is literally what was written in the solution I was given. How to write the solution properly ? Ker(f) = (z, -2z, z) and dim(Ker) = 1 ?

I would prefer to write "Ker(f)= {(z, -2z, z)}"- the set of all such vectors. Or you could write that the kernel is the space of all multiples of (1, -2, 1).